Integration and its applications
Integration in this exam tests the definite integral as accumulation, the fundamental theorem, and geometric applications such as area between curves and volumes of revolution. Questions often combine an algebraic curve with a line and ask for the bounded region.
Core rules
- Fundamental theorem: if F'=f, then the integral from a to b of f(x) dx is F(b)-F(a).
- Use substitution to reverse the chain rule and integration by parts for products; split rational functions before integrating.
- Area between upper curve y=u(x) and lower curve y=l(x) from a to b is the integral of u-l. Find intersections to set limits.
- Volume around the x-axis: V = π times the integral of y^2 dx. For a washer, integrate outer radius squared minus inner radius squared.
- Signed area matters: an integral can be negative, while geometric area is nonnegative. Split at zeros.
Worked example
Find the area between y=x^2 and y=2x. Set x^2=2x, so x(x-2)=0 and the curves meet at x=0 and x=2. On 0<x<2, 2x is above x^2. Thus area = integral from 0 to 2 of (2x-x^2) dx = [x^2 - x^3/3] from 0 to 2 = 4 - 8/3 = 4/3. A sketch confirms the upper-lower order and that no extra intersection lies in the interval.
Traps
Do not integrate y instead of y^2 for a volume, and do not forget the factor π. Reversing upper and lower curves gives a negative value. When substituting, change the limits or return to x before evaluating. For regions crossing the axis, integrate absolute value or split the interval; a single signed integral can cancel area.